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Can somebody show me how to prove, the statement in the title?

For given set, $A$, exist a set $\{(X,Y): X\subset Y\subset A \}$

Thanks!

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Let $X=Y=A$, and you just have $A\times A$. If $\subset$ is supposed to denote proper containment, then this is false for a set with 1 or 0 elements. –  rschwieb Oct 26 '12 at 12:28

1 Answer 1

up vote 2 down vote accepted

To show existence of a set $\{(X,Y): X\subset Y\subset A \}$ for an arbitrary set you can apply the Axiom schema of specification which says that "Every subclass of a set that is defined by a predicate is itself a set." To do this you would like to replace $\phi$ and $A'$ in the following schema:

$$ \forall w_1 , \dots, w_n \forall A' \exists B \forall x ( x \in B \iff [x \in A' \land \phi (x,w_1, \dots , w_n, A')])$$

Let $A'$ in the formula be the set of all pairs $(X,Y)$ with $X, Y$ sets and let $\phi (x, A') = (x=(X,Y)) \land (X \subset Y \subset A)$. Then $B$ is the set $\{(X,Y): X\subset Y\subset A \}$.

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I never understand what set theorists mean when they write something, but doesn't $\{(X,Y) : X \subset Y \subset A\}$ denote the the set of all pairs $(X,Y)$ of subsets of $A$ such that $X \subset Y$? I don't see why it suffices to write down a member of that collection and wrap it into braces. –  commenter Oct 26 '12 at 16:21
    
@commenter Thanks for noticing. You are absolutely right. I will try to fix this embarrassing mistake: I twisted its meaning inadvertently into "for a given set $A$ .... is non-empty". Funny that I even managed to BS the resident "set theorist to-be". (he will of course blame it on me) –  Matt N. Oct 26 '12 at 17:13
    
Dear @17SI.34SA I had to correct my answer because I misread your question. I'm sorry for any confusion I might have caused. It should be correct now. –  Matt N. Oct 26 '12 at 17:37
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Thanks for confirming and updating your answer. The resident ... preferred to sneak out by the back exit :-) –  commenter Oct 26 '12 at 21:23

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